Problem: $g(t) = 5t^{2}+4t+3(f(t))$ $f(t) = 3t^{3}+3t^{2}+6t+1$ $ g(f(0)) = {?} $
First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 3(0^{3})+3(0^{2})+(6)(0)+1$ $f(0) = 1$ Now we know that $f(0) = 1$ . Let's solve for $g(f(0))$ , which is $g(1)$ $g(1) = 5(1^{2})+(4)(1)+3(f(1))$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = 3(1^{3})+3(1^{2})+(6)(1)+1$ $f(1) = 13$ That means $g(1) = 5(1^{2})+(4)(1)+(3)(13)$ $g(1) = 48$